# Proof for $^{t}(A \cdot B) = ^{t}B \cdot ^{t}A$?

What's a simple proof for $^{t}(A \cdot B) = ^{t}B \cdot ^{t}A$ with $A \in M_{m,n}, B \in M_{n, p}$ ?

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Is that transposition? – J. M. Dec 30 '11 at 15:28
And is the $\cdot$ matrix multiplication? – Henning Makholm Dec 30 '11 at 15:29
Yes to both your questions – Mats Dec 30 '11 at 15:31
$(AB)^T = B^T A^T$ -- matrices multiply by juxtaposition. – Henning Makholm Dec 30 '11 at 15:47

The $(i,j)$ entry of $(AB)^T$ is the $(j,i)$ entry of $AB$, $\sum \limits_k a_{jk}b_{ki}$. The $(i,j)$ entry of $B^TA^T$ is $\sum \limits_k b_{ki}a_{jk}$ (the $i$th row of $B^T$ is the $i$th column of $B$, the $j$th column of $A^T$ is the $j$th row of $A$).

Just a tip, you should remember and understand that the $(i,j)$ entry of multiplying $AB$ is $\sum \limits_k a_{ik}b_{kj}$. As $k$ advances, you are going across row $i$ of $A$ and down column $j$ of $B$.

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Unfold the definitions of matrix multiplication and transposition in both cases. Notice that each element of the matrix on the two sides of the equals signs end up being the same sum of products of elements of $A$ and $B$.

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Could you give a worked out example, I mess up with the indexes. – Mats Dec 30 '11 at 15:38

For what's it's worth:

$$\overbrace{\left[\matrix{a_{11}&a_{12}&a_{13} \cr \color{maroon}{a_{21}}&\color{maroon}{a_{22}}&\color{maroon}{a_{23}} \cr a_{31}&a_{32}&a_{33} \cr }\right]}^A \overbrace{ \left[\matrix{b_{11}&b_{12}&\color{maroon}{b_{13}} \cr b_{21}&b_{22}&\color{maroon}{b_{23}} \cr b_{31}&b_{32}&\color{maroon}{b_{33}} \cr }\right]}^B = \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } \cr \cdot & \cdot & \cdot \cr }\right]}_{AB}$$ $$\phantom{ \overbrace{\left[\matrix{a_{11}&a_{12}&a_{13} \cr \color{maroon}{a_{21}}&\color{maroon}{a_{22}}&\color{maroon}{a_{23}} \cr a_{31}&a_{32}&a_{33} \cr }\right]}^A \overbrace{ \left[\matrix{b_{11}&b_{12}&\color{maroon}{b_{13}} \cr b_{21}&b_{22}&\color{maroon}{b_{23}} \cr b_{31}&b_{32}&\color{maroon}{b_{33}} \cr }\right]}^B=} \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \cdot \cr \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } & \cdot \cr }\right]}_{[AB]^T}$$ $$\underbrace{ \left[\matrix{b_{11}&b_{21}&b_{31} \cr {b_{12}}&{b_{22}}&{b_{32}} \cr \color{maroon}{b_{13}}&\color{maroon}{b_{23}}&\color{maroon}{b_{33}} \cr }\right]}_{B^T} \underbrace{ \left[\matrix{a_{11}&\color{maroon}{a_{21}}& {a_{31}} \cr a_{12}&\color{maroon}{a_{22}}& {a_{32}} \cr a_{13}&\color{maroon}{a_{23}}& {a_{33}} \cr }\right]}_{A^T} = \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \cdot \cr \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } & \cdot \cr }\right]}_{B^TA^T}$$

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Let $r_1,.., r_n$ be the rows of $A$ and $c_1,..,c_n$ be the columns of $B$.

The $ij$ entry in $(AB)^T$ is the $ji$ entry in $AB$, which is the dot product $r_j \cdot c_i$.

The $ij$ entry in $B^TA^T$ is the dot product between the $i$th row of $B^T$ and $j$th column of $A^T$, thus the dot product between the $i$th column of $B$ and $j$th row of $A$, hence $c_i \cdot r_j$.

Since

$$r_j\cdot c_i =c_i \cdot r_j$$

we get

$$(AB)^T=B^TA^T \,.$$

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